When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)
Solve -3c - 3c = 3c + 8z - 1 for c in terms of z.
| 2\(\frac{1}{5}\)z - 1\(\frac{2}{5}\) | |
| -1\(\frac{5}{6}\)z + \(\frac{1}{6}\) | |
| -z + 1 | |
| -3\(\frac{2}{3}\)z + 1\(\frac{2}{3}\) |